Alyosha
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I think problem 3 is asking for a strategy that skews the probability in favor of a high positive score, even though the expected value is still zero. For example: if you get the first question right, then skip the next question. If you get the first question wrong, then guess on the next question. Then you have a 75% chance of having a score of +1 after 2 questions (and 25% chance of having a score of -3.) Gamble! Probably there is some strategy with a high probability of some moderately positive score in exchange for a very low probability of some very very negative score.
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Here's a fun integral that I had to solve recently. Integral -inf to inf exp(-a^2 x^2)/(1 + x^2) dx Interested to see what approaches you take. (You may assume a is positive.)
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OmnipotentEntity wrote:
Here's a fun integral that I had to solve recently. Integral -inf to inf exp(-a^2 x^2)/(1 + x^2) dx Interested to see what approaches you take. (You may assume a is positive.)
Complex analysis is a glaring weakness in my math background. I'm not especially bad at it, I just tend to forget it readily. Frankly, I've never been impressed that the big draw is supposed to be that you can calculate certain integrals over the real line. I really should get into the deeper stuff like conformal mappings and such. Anyway, I jogged my memory by checking out Wikipedia's page on contour integrals. Example 2 there corresponds to your problem statement with t = i*a^2 (I promise I didn't peek at the answer). I then hopped over to the page on residues for help computing the residue of the pole at z=i. It's apparent from the numerator of the expression that the function goes to zero very quickly as |z| goes to infinity, so I computed the residue with a limit, imagined a contour along the real line and around that pole, and then took the real integral to be equal to the contour integral. My answer agrees with both Wikipedia's and Wolfram Alpha's.
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There are at least two other ways to solve this integral other than contour integration. Though I didn't realize that the answer was... just sitting on wikipedia for me to lift it. That would have been nice to know.
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Bobo the King wrote:
Anyway, I jogged my memory by checking out Wikipedia's page on contour integrals. Example 2 there corresponds to your problem statement with t = i*a^2
Example 2 with t = i*a^2 gives the integral of e^(-a^2 z)/(z^2+1).[1] OmnipotentEntity asked for the integral of e^(-a^2 z^2)/(z^2+1). The method described in Example 2 doesn't seem to work for e^(-a^2 z^2)/(z^2+1), primarily because the arc integral can't be bounded reasonably (values of z near pure imaginary numbers give values far from zero in the negative direction when squared; this gives a very large number for e^(-a^2 z^2)). In fact, WolframAlpha claims (citation) that the answer is not pi*e^(-a^2) (the expected answer being the contour integral around z=i), but actually pi*e^(-a^2)*(1-erf(a)), where erf is the error function. I wouldn't know how to prove this. [1] Actually, this isn't even valid. In Example 2, t specifically has to be a real number for the argument that the arc integral goes to 0 to hold.
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FractalFusion wrote:
Bobo the King wrote:
Anyway, I jogged my memory by checking out Wikipedia's page on contour integrals. Example 2 there corresponds to your problem statement with t = i*a^2
Example 2 with t = i*a^2 gives the integral of e^(-a^2 z)/(z^2+1).[1] OmnipotentEntity asked for the integral of e^(-a^2 z^2)/(z^2+1). The method described in Example 2 doesn't seem to work for e^(-a^2 z^2)/(z^2+1), primarily because the arc integral can't be bounded reasonably (values of z near pure imaginary numbers give values far from zero in the negative direction when squared; this gives a very large number for e^(-a^2 z^2)). In fact, WolframAlpha claims (citation) that the answer is not pi*e^(-a^2) (the expected answer being the contour integral around z=i), but actually pi*e^(-a^2)*(1-erf(a)), where erf is the error function. I wouldn't know how to prove this. [1] Actually, this isn't even valid. In Example 2, t specifically has to be a real number for the argument that the arc integral goes to 0 to hold.
Derp derp derp. Sloppy math on my part yet again...
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Here's the method I used to solve the integral: f(a) = integral e^(-a^2 x^2)/(x^2+1) dx f'(a) = integral -2a x^2 e^(-a^2 x^2)/(x^2+1) dx = integral -2a (x^2 + 1 - 1) e^(-a^2 x^2)/(x^2+1) dx = integral -2a e^(-a^2 x^2) dx + integral 2a e^(-a^2 x^2)/(x^2+1) dx f'(a) = -2sqrt(pi) + 2a f(a) This is just a linear non-homogenous first-order ODE. With solution: f(a) = pi e^(a^2) (-erf(a) + C) Because f(0) = pi we get C = 1 and 1 - erf(a) = erfc(a) so f(a) = pi e^(a^2) erfc(a)
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What is the result of the infinite sum: 3/4 + 3/16 + 3/64 + 3/256 + 3/1024 + ... Does the following infinite sum converge or diverge? 1/10 + 1/20 + 1/30 + 1/40 + 1/50 + 1/60 + ...
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Warp wrote:
What is the result of the infinite sum: 3/4 + 3/16 + 3/64 + 3/256 + 3/1024 + ... Does the following infinite sum converge or diverge? 1/10 + 1/20 + 1/30 + 1/40 + 1/50 + 1/60 + ...
The first is a geometric series with a = 3/4 and r = 1/4 The sum is a/(1-r) = 3/4 / (1 - 1/4) = 1 The second can be written as 1/10 ( 1 + 1/2 + 1/3 + 1/4 + ... ) The series in the brackets is the harmonic series, which is famously divergent.
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This was a problem from my Calc 2 course in 2002: If you take the harmonic series and remove only the terms that contain a 9 will the series converge or diverge? Prove it. If you want to cheat, this series is known as the Kempner series.
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OmnipotentEntity wrote:
If you take the harmonic series and remove only the terms that contain a 9 will the series converge or diverge? Prove it.
The series converges. Proof: Among all the positive integers that don't have the digit 9, there are 8 1-digit numbers, 8*9 2-digit numbers, 8*9*9 3-digit numbers, ... , 8*9k-1 k-digit numbers, and so on. Therefore the series is bounded above by 8(1)+8*9(1/10)+8*9*9(1/100)+...+8*9k-1(1/10k-1)+... = 8(1+9/10+(9/10)2+...+(9/10)k-1+...) = 8(1/(1-9/10)) = 8*10 = 80, and the summation terms are all positive and go to 0. Therefore it converges. Same argument if you exclude terms with the digit j, j something other than 9. (For j=0, the 8's in the proof above become 9's.)
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If this seems counter-intuitive: In the long run, most numbers have at least a digit '9'. As the number of digits goes to infinity, the percentage of those numbers goes to 100%.
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The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, 21, ... The Lucas numbers are: 2, 1, 3, 4, 7, 11, 18, 29, ... If you take the ratio between a Lucas number and the correspondent Fibonacci number, this ratio quickly converges to a specific value, the farther in the sequences you take this pair of numbers. This value seems to be approximately 1.38196601125. What value is this? (Obviously I'm not looking for an approximation but, if possible, the exact value.)
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Warp wrote:
The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, 21, ... The Lucas numbers are: 2, 1, 3, 4, 7, 11, 18, 29, ... If you take the ratio between a Lucas number and the correspondent Fibonacci number, this ratio quickly converges to a specific value, the farther in the sequences you take this pair of numbers. This value seems to be approximately 1.38196601125. What value is this? (Obviously I'm not looking for an approximation but, if possible, the exact value.)
That number is nowhere close to the final value. Starting from L(n+1) = F(n) + F(n+2) F(n+2) = F(n+1) + F(n) lim (n -> inf) F(n)/F(n+1) = [sqrt(5)-1]/2 We get: L(n+1) = F(n+1) + 2*F(n) => r(n+1) = L(n+1)/F(n+1) = 1 + 2*F(n)/F(n+1) => lim (n -> inf) r(n+1) = 1 + 2*[sqrt(5)-1]/2 = sqrt(5) So your ratio is sqrt(5) Edit: ah, I see what is going on: you are actually interested in L(n)/F(n+1) (you are missing F(0) in your list, making you of by one). Using the same identities in a slightly different manner: L(n+1) = 2*F(n+2) - F(n+1) => r(n+1) = L(n+1)/F(n+2) = 2 - F(n+1)/F(n+2) => lim (n -> inf) r(n+1) = 2 - [sqrt(5)-1]/2 = [5 - sqrt(5)]/2
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That's a good one Warp. Let's talk about matrix eigenvalue decomposition! So it is possible to represent both the Fibonnaci and Lucas recurrence relation as the same matrix with different initial conditions. This is to say, if you have a state vector (Fn-1, Fn) you can multiply it by a matrix to get the next state vector (Fn, Fn + 1) Our initial vector for Fibonacci numbers is (0, 1). For Lucas numbers it's (2, 1). Note: It seems that the limit you're computing is Ln / Fn+1 according to the canonical numbering. So the formula for the vector containing the (n-1)th and nth Fibonacci number is (0, 1) * ( ( 0, 1 ), ( 1, 1) )n But computing the nth power of a matrix is hard. So we need to find a better way. To begin, the matrix is simply:
A = ( ( 0, 1 ),
      ( 1, 1) )
Next, we calculate the eigenvalues of this matrix: ( ( 0-y, 1 ), ( 1, 1-y) ) (1-y)(-y) - 1 = 0 y2 - y - 1 = 0 y1 = (1 + sqrt(5))/2 y2 = (1 - sqrt(5))/2 (Note: this is the underlying mathematical reason that both the Fibonacci and Lucas sequence ratios approach the golden ratio as n goes to infinity.) So now that we have the eigenvalues we want to change the representation of the matrix from A into Q Y Q-1, where Y is just the matrix with the eigenvalues down the diagonal. We want to do this because then we can raise this to the nth power easily. (An is hard to calculate because it means lots of matrix multiplication. On the other hand, (Q Y Q-1)n is easy because the Q-1 and Q cancel and we're left with Q Yn Q-1. Because Y is a diagonal matrix, it is simple to calculate the nth power of, you just compute the nth power of each diagonal member.) It can be shown with a proof that Q must be the eigenvectors of A, but I'll omit that for brevity. Maybe this can be my challenge to you. So to calculate the eigenvectors of A we just solve the system (A - yI) v = 0 for v. I'll spare the calculations because they're pretty much trivial we get: v1 = (-y2, 1) v2 = (-y1, 1) So Q becomes: ( ( -y2, -y1 ), ( 1, 1) ) And Q-1 is computed just by taking the inverse of Q: ( ( 1/sqrt(5), (1 + sqrt(5))/(2 sqrt(5)) ), ( -(1/sqrt(5)), (-1 + sqrt(5))/(2 sqrt(5)) ) ) Now we can use this to find an explicit formula for the nth Fibonacci (and Lucas) number. It's just: (0, 1) Q Yn Q-1 And because we just care about a single value we can simplify it to a single formula: (-(1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n)/sqrt(5) Which is the famous explicit formula for the Fibonacci sequence. Similarly you can use the initial vector for the Lucas sequence to find the explicit formula for that. It is: -2 + (1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n Because the limit you're looking for is Ln / Fn+1 we have: (-2 + (1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n) / ((-(1/2 (1 - sqrt(5)))^(n+1) + (1/2 (1 + sqrt(5)))^(n+1))/sqrt(5)) Which is messy looking but as n approaches infinity is: 1/2 ( 5 - sqrt(5) ) For Ln / Fn the formula is: (-2 + (1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n) / ((-(1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n)/sqrt(5)) And the limit is simply: sqrt(5) Which is a lot nicer. :)
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I saw this on a t-shirt: Is that really so?
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Well, you can simplify this a lot, but yes. cos(pi/3) = 1/2 ln(e^(1/3)) = 1/3 int(z^2 dz) from 1 to 3^(1/3) is 2/3. So this boils down to 2/3 * 1/2 = 1/3, which is true.
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Warp wrote:
I saw this on a t-shirt: Is that really so?
Yes, but it looks quite peculiar. It's off in a way that makes you think it was composed by someone who haphazardly combined pieces but didn't understand what they were looking at, and they were lucky it ended up working. Like, why isn't 3π/9 reduced to π/3? That's elementary school arithmetic, it's just strange and silly when you're not doing something esoteric like trying to use every digit. Why is the cosine positioned so ambiguously? It looks weird and makes you wonder if it's intended to be part of the integral or not. Yeah, it's to the right of the dz, so in the most formal/technical sense it's not part of the integral, but any sane person would move the cosine in front of the integral, or otherwise visually separate them somehow. (It doesn't actually matter either way in this case, but it looks weird.) It would also be completely standard to omit all of the parentheses as shown in the original equation, and failing to do so kind of makes the whole thing resemble typing in ASCII or something, even though it's clearly typeset. Lastly, there should be space between z^2 and dz. A lot of this stuff might seem nitpicky, but math is in many ways the art of communicating consistently and unambiguously, which the shirt you sighted fails to do. Someone that knew what they were doing would probably write the equation like this, perhaps with a different choice of multiplication symbol:
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If you took the time to talk to the person wearing the shirt they'd have probably told you that it's a limerick. The Integral z squared dz From one to the cube root of three times the cosine of three pi over nine is the log of the cube root of e.
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Nice catch. Here's another mathematical limerick. (12 + 144 + 20 + 3x4^0.5 ) / 7 + 5x11 = 9^2 A dozen a gross and a score plus three times the square root of four divided by seven plus five times eleven is nine squared and not a bit more.
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9=9 These limericks are pretty fine Now look here is the number nine It's pretty cool I am no fool Turns out it equals nine
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There once was a man from Peru Whose limericks stopped at line two There once was a man from Verdun Along those lines, has anyone heard the limerick about the guy from Nero?
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Sadly, that limerick (the first one mentioned) doesn't even rhyme properly as far as us Brits are concerned.
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Then you can replace zed with any of the following: b, c, g, p, t, v d can't be used because it's the differential, e can't be used because we're using it as a constant. t might work better because it's a more often used variable, and z generally denotes a complex number, which isn't strictly required in this context.
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FractalFusion wrote:
OmnipotentEntity wrote:
If you take the harmonic series and remove only the terms that contain a 9 will the series converge or diverge? Prove it.
The series converges. Proof: Among all the positive integers that don't have the digit 9, there are 8 1-digit numbers, 8*9 2-digit numbers, 8*9*9 3-digit numbers, ... , 8*9k-1 k-digit numbers, and so on. Therefore the series is bounded above by 8(1)+8*9(1/10)+8*9*9(1/100)+...+8*9k-1(1/10k-1)+... = 8(1+9/10+(9/10)2+...+(9/10)k-1+...) = 8(1/(1-9/10)) = 8*10 = 80, and the summation terms are all positive and go to 0. Therefore it converges. Same argument if you exclude terms with the digit j, j something other than 9. (For j=0, the 8's in the proof above become 9's.)
It's interesting to think about it, as I'm pretty sure if you remove any sequence of numbers, it will also converge. I wonder what the limit will be if, for instance, we remove only the terms that have a ''1234567890''.
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